Find Intersection point between two arrays

In the intersection part am not getting it. Error in fzero. What function to use to plot the intersection

format long

clc

clear all

% Target Characteristics pre-defined parameters

xt = 2.3/1000; % target (x) in m

yt = 2.3/1000; % target (y) in m

p = 50/100; % percent target

vis = 23; % Visibility in km target

er = 38; % LRF

dref = 500/1000; % LRF distance in m

pref = 85/100; % LRF

T = 90/100; % LRF

wave= 1.543;

%Atmospheric transmission

x= 0:1:10;

b=(0.55/wave);

a=(3.912/vis);

c=0.5*(vis^(1/3));

dbc=b^c;

atmtrans = a*dbc ; % Atmospheric transmission equation

% beam area

angle= 0.6/1000/2; % div angle in mrad

B=[];

sc=[];

A=[];

deno=[];

f1 = [];

f2 = [];

f = [];

xsol = [];

ysol =[];

% distance range in km

for d= 0:1:10

R = tan(angle).*d ; % radius vary with distance in mm

beamArea = (pi.*R.*R) ; % beam area in km2

%target area

targetArea=(xt*yt); % in km2

sc= [sc ((beamArea)./targetArea)]; %ratio between the Atarget and Abeam

% Attenuation

A = [A exp(-2*atmtrans.*d)]; %in km

% Sensitivity

pratio=pref/p

dratio=(d.*d)/(dref^2);

ex=exp(atmtrans *dref*-2);

er10=-10^(er/10);

deno=[T.*sc];

B = [(pratio.*dratio).*((ex*er10)./deno) ];

end

% Plotting curves initially

plot(x, A, x, B)

% Finding the intersection point

for x= 0:1:10

f1 = @(x) [ A exp(-2*atmtrans.*d) ];

f2 = @(x) [(pratio.*dratio).*((ex*er10)./deno) ];

f = @(x) [f1(x) - f2(x)];

x0 = 0.5; %initial guess value (intersection is somewhere near x = 0.5)

xsol = fzero(f, x0)

ysol = f1(xsol)

end

% Marking the intersection point

plot(x, A, x, B), hold on

plot(xsol,ysol, 'mo', 'MarkerSize', 14), hold on

ylim([0 1])

xlabel("Range in km")

ylabel("Distance in km")

title("Range Preformance")

legend('Atmospheric Attenuation','Sensitivity','Intersection')

hold off

[Star Strider]

Neither ‘f1’ nor ‘f2’ are functions of ‘x’, and the vectors that comprise them are not the same sizes. If they were, and all the elements were finite, I would have suggested using interp1 instead.

[Anfal AlShehhi]

How to use this func

[Star Strider]

With respect to interp1, here is an example —

x = [0 10];
Line1 = [x(:) ones(2,1)] * randn(2,1)

Line1 = 2×1

-1.0596 11.6990

Line2 = [x(:) ones(2,1)] * randn(2,1)

Line2 = 2×1

0.0584 9.2103

xint = interp1(Line1 - Line2, x, 0)                             % X-Intersection Value

xint = 3.0999

yint = interp1(x, Line1, xint)                                  % Y-Intersection Value

yint = 2.8954

figure
plot(x, Line1)
hold on
plot(x, Line2)
plot(xint, yint, 'rs')
hold off
grid
legend('Line1','Line2','Intersection', 'Location','best')

Providing an intersection exists in the region-of-interest, this approach will work.

.

[Mathieu NOE]

hello @Anfal AlShehhi

you can streamline and vectorize your code

no need for for loop

x and d are both used for the same job can be surce of trouble.

I guess there is something wrong with the equations as A and B curves are so off that no risks of crossing at x = 0.5

like this line , th end seems a bit suspect to me :

ex=exp(atmtrans *dref*-2); % how to interpret the -2 ?

ex=exp(-2*atmtrans *dref); ??

clc
clearvars

% Target Characteristics pre-defined parameters
xt   = 2.3/1000;         % target (x) in m
yt   = 2.3/1000;         % target (y) in m
p    = 50/100;           % percent target
vis  = 23;               % Visibility in km target
er   = 38;               % LRF 
dref = 500/1000;         % LRF distance in m
pref = 85/100;           % LRF 
T    = 90/100;           % LRF 
wave=  1.543;
%Atmospheric transmission
d= 0.1:0.1:10 ; % distance range in km
                % do not start at zero (otherwise division by zero occurs in B = [(pratio.*dratio).*((ex*er10)./deno) ];

b=(0.55/wave);
a=(3.912/vis);
c=0.5*(vis^(1/3));
dbc=b^c;
atmtrans = a*dbc  ;            % Atmospheric transmission equation
 
% beam area
angle= 0.6/1000/2;             % div angle in mrad
 
R = tan(angle).*d ;  % radius vary with distance in mm
beamArea = (pi.*R.*R) ;   % beam area in km2
%target area
targetArea=(xt*yt); % in km2
sc = (beamArea./targetArea);      %ratio between the Atarget and Abeam 
 
% Attenuation 
A = exp(-2*atmtrans.*d); %in km
 
% Sensitivity
pratio = pref/p;
dratio=(d.*d)/(dref^2);
ex=exp(atmtrans *dref*(-2)); % ex=exp(atmtrans *dref*(-2)) ??
% ex=exp(atmtrans *dref.^(-2)); % ex=exp(atmtrans *dref*(-2)) ??
er10=-10^(er/10);
deno=[T.*sc];
B = [(pratio.*dratio).*((ex*er10)./deno) ];


% Plotting   curves initially
plot(d, A, d, B)

Answer 1

[Nipun]

Hi Anfal,

I understand that you are trying to find the intersection between two arrays that represent plots on 2D surface.

I recommend using curve intersections by NS to find the intersection of two curves. Download and unzip InterX.m and the following code after plotting A and B:

 p = InterX([d;A],[d;B]);
 hold on
 for i=1:length(p)
    plot(p(1,i), p(2,i), 'rs');
 end

All points in p correspond to intersection points. If p is empty, then there are no intersection points. Hope this helps.

Regards,

Nipun